# Generate Parentheses

To explain the problem of "Generate Parentheses," let's start from the very basic principles and then build up to the complexity of the problem.

### Basic Concepts:

1. Parentheses: Parentheses are symbols used in writing to include material that clarifies or is a side note. The symbols are "(" for the opening parenthesis and ")" for the closing parenthesis.

2. Balanced Parentheses: A string of parentheses is balanced if every opening parenthesis has a corresponding closing parenthesis and the pairs of parentheses are properly nested. For example, "()" and "(())" are balanced, but "(()" and "())(" are not.

### The Problem Statement:

The "Generate Parentheses" problem asks you to generate all combinations of well-formed parentheses for a given number n, where n is the number of pairs of parentheses.

### Understanding the Problem with an Example:

For n = 3, the goal is to generate all combinations of three pairs of balanced parentheses. These combinations would include:

• "((()))"
• "(()())"
• "()(())"
• "()()()"
• "((())())"

### Principles for Generating Parentheses:

1. Recursion: This problem is naturally suited for a recursive approach. Each step of the recursion adds either an opening or a closing parenthesis until all parentheses are used.

2. Maintaining Balance: To ensure the parentheses remain balanced, we can only add a closing parenthesis if there are fewer closing parentheses than opening ones at any point in the string being constructed.

3. Base Condition: When the total number of opening and closing parentheses used equals 2n, a valid combination has been formed.

### Algorithmic Approach:

• Add an opening parenthesis if the number of opening parentheses is less than n.
• Add a closing parenthesis if the number of closing parentheses is less than the number of opening parentheses.
3. When the length of the string is 2n, add the string to the list of results.

### Pseudocode:

function generateParenthesis(n):
result = []
function backtrack(S = '', left = 0, right = 0):
if len(S) == 2 * n:
result.append(S)
return
if left < n:
backtrack(S + '(', left + 1, right)
if right < left:
backtrack(S + ')', left, right + 1)

backtrack()
return result

In this pseudocode:

• S is the current string being formed.
• left and right keep track of the number of opening and closing parentheses used.
• The backtrack function is called recursively to build the string.

### Complexity:

• Time Complexity: O(4^n / sqrt(n)). This is a rough estimate due to the complexity of the Catalan number which is involved in counting the number of valid parentheses.
• Space Complexity: O(4^n / sqrt(n)) for storing all valid combinations, and O(n) for the recursion stack.

This approach comprehensively covers the problem from its basic principles to the detailed algorithm, ensuring clarity and logical reasoning throughout.